Find the 3rd term of \((\frac{x}{2} - 1)^{8}\) in descending order of x.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option B
Explanation:
\((\frac{x}{2} - 1)^{8} = ^{8}C_{8}(\frac{x}{2})^{8}(-1)^{0} + ^{8}C_{7}(\frac{x}{2})^{7}(-1)^{1} + ^{8}C_{6}(\frac{x}{2})^{6}(-1)^{2} + ...\)
\(\text{The third term in the expansion =} ^{8}C_{6}(\frac{x}{2})^{6}(-1)^{2}\)
= \(\frac{8!}{6!2!}(\frac{x^{6}}{64})(1) \)
= \(28 \times \frac{x^{6}}{64} = \frac{7x^{6}}{16}\)
\((\frac{x}{2} - 1)^{8} = ^{8}C_{8}(\frac{x}{2})^{8}(-1)^{0} + ^{8}C_{7}(\frac{x}{2})^{7}(-1)^{1} + ^{8}C_{6}(\frac{x}{2})^{6}(-1)^{2} + ...\)
\(\text{The third term in the expansion =} ^{8}C_{6}(\frac{x}{2})^{6}(-1)^{2}\)
= \(\frac{8!}{6!2!}(\frac{x^{6}}{64})(1) \)
= \(28 \times \frac{x^{6}}{64} = \frac{7x^{6}}{16}\)